Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

<image>

You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3F3F3F3F;
const int MOD = 1e9 + 7;
const int MAX = 2e5 + 5;
int Y, A, B, N, Q;
pair<int, int> segs[MAX];
int pref[MAX];
inline int query(int l, int r) { return pref[r] - pref[l - 1]; }
inline double findX(int segPt, pair<int, int> pt, double k) {
  return pt.first + k * (segPt - pt.first);
}
inline int bSearch(pair<int, int> pt, int x, double k) {
  int lo = 1, hi = N + 1, mid;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    if (x < findX(segs[mid].second, pt, k)) {
      hi = mid;
    } else {
      lo = mid + 1;
    }
  }
  return lo;
}
inline bool contain(pair<int, int> seg, int x, pair<int, int> pt, double k) {
  return x > findX(seg.first, pt, k) && x < findX(seg.second, pt, k);
}
int main() {
  scanf("%d%d%d", &Y, &A, &B);
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &segs[i].first, &segs[i].second);
    pref[i] = pref[i - 1] + segs[i].second - segs[i].first;
  }
  scanf("%d", &Q);
  pair<int, int> pt;
  for (int i = 1; i <= Q; i++) {
    scanf("%d%d", &pt.first, &pt.second);
    double k = (double)(pt.second - Y) / pt.second;
    double ans = query(1, N) * k;
    int l = bSearch(pt, A, k), r = bSearch(pt, B, k);
    if (contain(segs[l], A, pt, k)) {
      ans += findX(segs[l].second, pt, k) - A;
      l++;
    }
    if (contain(segs[r], B, pt, k)) {
      ans += B - findX(segs[r].first, pt, k);
    }
    ans -= query(1, l - 1) * k, ans -= query(r, N) * k;
    printf("%.15lf\n", ans);
  }
  return 0;
}